$I=(2,X)$ and $J=(3,X)$

Question

For $I,J\in\mathbb{Z}[X]$ with $I=(2,X)$ and $J=(3,X)$, why do we have $\{i\cdot j:i\in I,j\in J\}\neq I\cdot J=(6,X)$? And why isn't $\{i\cdot j:i\in I,j\in J\}$ an ideal in $\mathbb{Z}[X]$?

Answer 1

Consider the element $6+X$. If $6+X \in \{ i \cdot j : i \in I,\ j \in J \}$, then there exist $a,b,c,d \in \mathbb{Z}$ such that $$6+X = (2a+bX)(3c+dX) = 6ac+(2ad+3bc)X+bdX^2$$ (There certainly can't be any quadratic or higher terms in the factors since that would force $6+X$ to have degree greater than $2$.)

Comparing coefficients gives $bd=0$, so $b=0$ or $d=0$. If $b=0$ then $2ad=1$, and if $d=0$ then $3bc = 1$; both of these are impossible since $2 \nmid 1$ and $3 \nmid 1$.

However, the fact that $6+X \in (6, X)$ is plain.

This proves that $IJ \ne \{ ij : i \in I,\ j \in J \}$. To see that $\{ ij : i \in I,\ j \in J \}$ is not an ideal, simply note that $6$ and $X$ are both elements but that $6+X$ is not.

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In general, given ideals $I,J$ in a ring $R$, we have $$IJ = \left\{ \sum_{k=0}^n i_kj_k : n \in \mathbb{N} \text{ and } i_k \in I,\ j_k \in J \text{ for all } k \le n \right\}$$ which is usually not equal to $\{ ij : i \in I,\ j \in J \}$.

Answer 2

$IJ$ can be defined as the smallest ideal which contains the products $ij$, where $i\in I, j\in J$. Unless one of $I,J$ is a principal ideal, the set of these products is not an ideal.